integrate(exp(exp(I*x)), [x, 0, 2*pi]) incorrectly 0

[ethankward]

>>> from sympy import *
>>> x = Symbol('x')
>>> integrate(exp(exp(I*x)), [x, 0, 2*pi])
0

The answer should be 2*pi.

[ethankward]

This returns an unevaluated integral in 1.3.

[jksuom]

There are many examples of the same behaviour among older issues. When the integrand is periodic, its antiderivative will usually also turn out to be periodic. It follows that if the integral over a period is not zero, the antiderivative will be discontinuous. That should be checked before applying the fundamental theorem of calculus (that only works with continuous antiderivatives).

[oscarbenjamin]

Is it possible to solve these by detecting periodic functions with discontinuous antiderivatives and then using a patched up antiderivative like expr_discont + delta * (x % period) where delta is the integral over the period?

Related issue #15730

[jksuom]

Something like that will do if the discontinuities are at n*period for integers n. But perhaps there should rather be floor(x/period) instead of x % period.

[oscarbenjamin]

As long as the discontinuities are periodic (which they will be if the antiderivative is periodic) then it should be possible to add a number of different terms like delta_i * floor((x-x_i)/period) where x_i is the location of the discontinuity and delta_i is the difference across the discontinuity.

[jksuom]

I think that would work. Only, one has to find all the x_i and delta_i. The idea has already been implemented in some cases involving atans.

[oscarbenjamin]

Are there already methods for detecting if an expression is periodic or continuous or for finding discontinuities?

[jksuom]

There is periodicity in calculus.util. Discontinuities are harder to find, especially, when branched multi-valued functions are involved.

[oscarbenjamin]

In this particular case the antiderivative is

In [5]: integrate(exp(exp(I*x)))                                                                                                               
Out[5]: 
     ⎛ ⅈ⋅x⎞
-ⅈ⋅Ei⎝ℯ   ⎠

Periodicity doesn't work presumably because it doesn't understand exp(I*x):

In [14]: from sympy.calculus.util import periodicity                                                                                           

In [15]: print(periodicity(exp(I*x), x))                                                                                                       
None

[Teut2711]

Hey, can I work on this issue? But I 'm still not good enough to do this alone since I have only fixed a single issue that too partly.

[oscarbenjamin]

I don't know what can be done for solving this kind of integral. I described something above but it's not simple because other features for finding discontinuities etc. don't exist. periodicity is already being worked on in #16068

[Teut2711]

well if you solve it by hand you use substitution e^ix = t but I don't know how does sympy use substitution .

[mohitacecode]

In the current master the following result is returned but it isn't getting simplified it is returning same expression is it the case when sympy don't have the solution for the problem it returned unsolved form?

>>> Integral(abs(sin(phi)*cos(phi)), (phi,pi,2*pi))
Integral(Abs(sin(phi)*cos(phi)), (phi, pi, 2*pi))
>>> _.doit()
Integral(Abs(sin(phi)*cos(phi)), (phi, pi, 2*pi))
>>> 

[moorepants]

is it the case when sympy don't have the solution for the problem it returned unsolved form?

Yes, this is the expected behavior: return the Integral if SymPy can solve it.